# 22.4.3: iii. Exercise Solutions

## Q1

a. $$CCl_4$$ is tetrahedral and therefore is a spherical top. $$CHCl_3 \text{ has } C_{3v}$$ symmetry and therefore is a symmetric top. $$CH_2Cl_2 \text{ has } C_{2v}$$ symmetry and therefore is an asymmetric top.
b. $$CCl_4$$ has such high symmetry that it will not exhibit pure rotational spectra. $$CHCl_3 \text{ and } CH_2Cl_2$$ will both exhibit pure rotation spectra.
c. $$NH_3$$ is a symmetric top (oblate). Use the given energy expression,|
$E = (A - B) K^2 + B J(J + 1),$
A = 6.20 $$cm^{-1}$$, B = 9.44$$cm^{-1}$$, selection rules $$\Delta J = \pm 1$$, and the fact that $$\vec{\mu_0}$$ lies along the figure axis such that $$\Delta K = 0$$, to give:
$\Delta E = 2B (J + 1) = 2B, 4B\text{, and } 6B (J = 0, 1 \text{, and } 2).$
So, lines are at 18.88$$cm^{-1}$$, 37.76$$cm^{-1}$$. and 56.64$$cm^{-1}$$.

## Q2

To convert between $$cm^{-1}$$ and energy, multiply by hc = $$6.62618x10^{-34}\text{ J sec })(2.997925x10^{10}\text{ cm sec}^{-1}) = 1.9865x10^{23} \text{ J cm }$$.
Let all quantities in $$cm^{-1}$$ be designated with a bar,
e.g. $$\bar{B_e}$$ = 1.78$$cm^{-1}$$.
a. \begin{align} hc\bar{B_e} &=& &\dfrac{\hbar^2}{2\mu R_e^2} \\ R_e &=& &\dfrac{\hbar}{\sqrt{2\mu hc\bar{B_e}}}, \\ \mu &=& &\dfrac{m_Bm_O}{m_B + m_O} = \dfrac{(11)(16)}{(11 + 16)}x1.66056x10^{-27} \text{kg} \\ &=& &1.0824x10^{26} \text{kg} \\ hc\bar{B_e} &=& &hc(1.78 \text{cm}^{-1} = 3.5359x10^{-23} \text{J} \\ R_e &=& & \dfrac{1.05459x10^{-34}\text{J sec}}{\sqrt{(2)1.0824x10^{-26}\text{kg}.3.35359x10^{-23}\text{ J}}} \\ R_e &=& &1.205x10^{-10}\text{m} = 1.205 Å \\ D_e &=& &\dfrac{4B_e^3}{\hbar \omega_e^2}, \bar{D_e} = \dfrac{4\bar{B_e}^3}{\bar{\omega_e}^2} = \dfrac{(4)(1.78 \text{cm}^{-1})^3}{(4)(66782.2 \text{cm}^{-1})} = 6.35x10^{-6}\text{cm}^{-1} \\ \omega_ex_e &=& &\dfrac{\hbar \omega_e^2}{4D_e^0}, \bar{\omega_ex_e} = \dfrac{\bar{\omega_e^2}}{4\bar{D_e^0}} = \dfrac{(1885 \text{cm}^{-1})^2}{(4)(66782.2 \text{cm}^{-1})} = 13.30 \text{cm}^{-1}.\\ D_0^0 &=& &D_e^0 - \dfrac{\hbar\omega_e}{2} + \dfrac{\hbar\omega_e x_e}{4}, \bar{D_0^0} = \bar{D_e^0} - \dfrac{\omega_e}{2} + \dfrac{\bar{\omega_ex_e}}{4} \\ &=& &66782.2 - \dfrac{1885}{2} + \dfrac{13.3}{4} \\ &=& &65843.0 \text{cm}^{-1}= 8.16 eV. \\ \alpha_e &=& & \dfrac{-6B_e^2}{\hbar\omega_e} + \dfrac{6\sqrt{B_e^3\hbar\omega_e x_e}}{\hbar\omega_e} \\ \bar{\alpha_e} &=& & \dfrac{-6\bar{B_e^2}}{\bar{\omega_e}} + \dfrac{6\sqrt{\bar{B_e^3}\bar{\omega_e x_e}}}{\bar{\omega_e}} \\ \bar{\alpha_e} &=& &\dfrac{(-6)(1.78)^2}{(1885)} + \dfrac{6\sqrt{(1.78)^3(13.3)}}{(1885)} = 0.0175 \text{cm}^{-1}. \\ B_0 &=& & B_e - \alpha_e \left( \dfrac{1}{2} \right), \bar{B_0} = \bar{B_e} - \bar{\alpha_e}\left( \dfrac{1}{2} \right) = 1.78 - \dfrac{0.0175}{2} \\ &=& & 1.77 \text{cm}^{-1} \\ B_1 &=& & B_e - \alpha_e\left( \dfrac{3}{2} \right), \bar{B_1} = \bar{B_e} - \bar{\alpha_e} \left( \dfrac{3}{2} \right) = 1.78 - 0.0175(1.5) \\ &=& & 1.75 \text{cm}^{-1} \end{align}
b. The molecule has a dipole moment and so it should have a pure rotational spectrum. In addition, the dipole moment should change with R and so it should have a vibration rotation spectrum.
The first three lines correspond to J $$=1 \rightarrow 0, J=2 \rightarrow 1, J=3 \rightarrow 2$$
\begin{align} E &=& & \hbar \omega_e\left( v + \dfrac{1}{2}\right) - \hbar \omega_ex_e \left( v + \dfrac{1}{2}\right)^2 + B_vJ(J + 1) - D_eJ^2(J + 1)^2 \\ \Delta E &=& &\hbar \omega_e - 2\hbar \omega_ex_e - B_0J)J + 1) + B_1J(J - 1) - 4D_eJ^3 \\ \bar{\Delta E} &=& & \bar{\omega_e} - 2\bar{\omega_e x_e} - \bar{B_0}J(J = 1) + \bar{B_1}J(J - 1) - 4\bar{D_e}J^3 \\ \bar{\Delta E} &=& & 1885 - 2(13.3) - 1.77J(J + 1) + 1.75J(J - 1) - 4(6.35x10^{-6})J^3 \\ &=& & 1858.4 - 1.77J(J + 1) + 1.75J(J - 1) - 2.54x10^{-5}J^3 \\ \bar{\Delta E}(J = 1) &=& & 1854.9 \text{cm}^{-1} \\ \bar{\Delta E}(J = 2) &=& & 1851.3 \text{cm}^{-1} \\ \bar{\Delta E}(J = 3) &=& & 1847.7 \text{cm}^{-1} \end{align}

## Q3

The $$C_2H_2Cl_2$$ molecule has a $$\sigma_h$$ plane of symmetry (plane of molecule) a $$C_2$$ axis ($$\bot$$ to plane), and inversion symmetry, this result in $$C_{2h}$$ symmetry. Using $$C_{2h}$$ symmetry labels the modes can be labeled as follows: $$\nu_1, \nu_2, \nu_3, \nu_4 \text{, and } \nu_5 \text{ are } a_g, \nu_6 \text{ and } \nu_7 \text{ are } a_u, \nu_8 \text{ is } b_g \text{, and } \nu_9, \nu_{10}, \nu_{11} \text{, and } \nu_{12} \text{ are } b_u.$$